Question

我的问题是，根据我的示例，我已经能够通过指针指向一个非常量变量，该指针指向一个常量变量。我的样本是这样的： -

int tobepointed = 10;
int tobepointed1 = 11;
int const *ptr = &tobepointed;
cout << "\nPointer points to the memory address: " << ptr;
cout << "\nPointer points to the value: " << *ptr;
tobepointed = 20;
cout << "\nPointer points to the memory address: " << ptr;
cout << "\nPointer points to the value: " << *ptr;
ptr = &tobepointed1;
cout << "\nPointer points to the memory address: " << ptr;
cout << "\nPointer points to the value: " << *ptr;

现在这段代码正常运行，没有任何编译或运行时错误。还要考虑指针* ptr。如果我像任何普通指针一样声明ptr： - int *ptr;

那么输出也是一样的，为什么我们需要'指向常数'的概念？

是的我同意'常量指针'的概念，但“概念指针”似乎毫无用处。

Answer 1

指向常量的指针应指向...常量int。您的程序会编译，因为您指的是tobepointed int。

如果你声明它const int，那么指向它（没有强制转换）的唯一方法就是使用const int*：

const int tobepointed = 10;
const int* ptr = &tobepointed;

如果你试图创建非常量指针：

const int tobepointed = 10;
int* ptr = &tobepointed;

编译器would complain：

错误：来自＆＃39; const int *＆＃39;的无效转换到＆＃39; int *＆＃39;

'指向常数'的用法是什么？

1 个答案: